Tạm thời chỉ nghĩ ra được cách này -_- 

Ta có : 

\(A=\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}\)

\(A=\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2014+2}{2014}\)

\(A=\frac{2015}{2015}-\frac{1}{2015}+\frac{2016}{2016}-\frac{1}{2016}+\frac{2014}{2014}+\frac{2}{2014}\)

\(A=1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{2}{2014}\)

\(A=\left(1+1+1\right)-\left(\frac{1}{2015}+\frac{1}{2016}-\frac{2}{2014}\right)\)

\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]\)

Lại có : 

\(\frac{1}{2015}< \frac{1}{2014}\)

\(\frac{1}{2016}< \frac{1}{2014}\)

\(\Rightarrow\)\(\frac{1}{2015}+\frac{1}{2016}< \frac{1}{2014}+\frac{1}{2014}\)

\(\Rightarrow\)\(\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)< 0\)

\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]>3\)

Vậy \(A>3\)

Chúc bạn học tốt ~ 

A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)

\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)

\(Vậy:A>B\)

Đúng nha Nguyễn Bình Minh

so sánh:

\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)  và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)

                                                             \(B=\frac{2013}{2014+2015+2016}+\frac{2014}{2014+2015+2016}+\frac{2015}{2014+2015+2016}\)

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)

          \(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)

          \(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)

\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}>\frac{2013+2014+2015}{2014+2015+2016}\)

Vậy: \(A>B\)

A = (n + 2015)(n + 2016) + n² + n

= (n + 2015)(n + 2015 + 1) + n(n + 1)

Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2

=> (n + 2015)(n + 2015 + 1) chia hết cho 2

      n(n + 1) chia hết cho 2

=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2

=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)

so sánh: \(A=\frac{2014}{2015}+\frac{2015}{2016}\)  và \(B=\frac{2014+2015}{2015+2016}\)

                                               \(\Rightarrow B=\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)

Ta có: \(\frac{2014}{2015}>\frac{2014}{2015+2016}\) vì \(2015<2015+2016\)

          \(\frac{2015}{2016}>\frac{2015}{2015+2016}\) vì \(2016<2015+2016\)

\(\Rightarrow\frac{2014}{2015}+\frac{2015}{2016}>\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)

\(\Rightarrow\frac{2014}{2015}+\frac{2015}{2016}>\frac{2014+2015}{2015+2016}\)

Vậy:     \(A>B\)

\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(1-\frac{1}{2015}\right)+\left(1-\frac{1}{2016}\right)+\left(1+\frac{2}{2014}\right)\)

                                   \(=3-\left(\frac{1}{2015}-\frac{1}{2016}+\frac{2}{2014}\right)\)

Dễ thấy \(\frac{1}{2015}-\frac{1}{2016}+\frac{2}{2014}>0\) vì \(\frac{1}{2015}>\frac{1}{2016}\)

Do đó \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}< 3\)

A = 2014*2015 + 2015/2016 + 2016/2014

A = (1 - 1/2015) + (1 - 1/2016) + (1 + 2/2014)

A = 3 + (2/2014 - 1/2015 - 1/2016)

A = 3 + (2*2015*2016 - 2014*2016 - 2014*2015) / (2014*2015*2016)

Đặt B = 2*2015*2016 - 2014*2016 - 2014*2015

Ta có: A = 3 + B/(2014*2015*2016)

Nhận xét: Từ các phép biến đổi trên ta thấy A là tổng của 3 với một phân số có mẫu số dương. Do vậy, để so sánh A với 3 ta chỉ cần so sánh B với 0.

B = 2*2015*2016 - 2014*2016 - 2014*2015

B = 2016(2*2015 - 2014) - 2014*2015

B = 2016(2*2015 - 2014) - 2014(2016 - 1)

B = 2016(2*2015 - 2014) - 2014*2016 + 2014

B = 2016(2*2015 - 2014 - 2014) + 2014

B = 2016(2*2015 - 2*2014) + 2014

B = 2*2016(2015 - 2014) + 2014

B = 2*2016 + 2014 > 0

Vậy A > 3 (Đáp số)